Chemical Principles: The Quest for Insight
Chemical Principles: The Quest for Insight
7th Edition
ISBN: 9781464183959
Author: Peter Atkins, Loretta Jones, Leroy Laverman
Publisher: W. H. Freeman
Question
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Chapter F, Problem I.26E

(a)

Interpretation Introduction

Interpretation:

Chemical equation, complete ionic equation and also the net ionic equation for the reaction between (NH4)3PO4 and CaCl2 has to be written.

(a)

Expert Solution
Check Mark

Explanation of Solution

The reaction given is between aqueous solutions of (NH4)3PO4 and CaCl2.  The reaction can be given as follows:

    2(NH4)3PO4(aq)+3CaCl2(aq)Ca3(PO4)2(s)+6NH4Cl(aq)

Aqueous solution of (NH4)3PO4 reacts with CaCl2 resulting in formation of Ca3(PO4)2 and NH4Cl.  The precipitate that is formed is Ca3(PO4)2.

The balanced chemical equation for the reaction is given as follows:

    2(NH4)3PO4(aq)+3CaCl2(aq)Ca3(PO4)2(s)+6NH4Cl(aq)

Complete ionic equation is written by showing the ions present in aqueous solution.  This is given as follows:

6NH4+(aq)+2PO43(aq)+3Ca2+(aq)+6Cl(aq)Ca3(PO4)2(s)+6NH4+(aq)+6Cl(aq)

In the complete ionic equation, the common ions that are present on both sides of the equation is known as spectator ions.  Thus in this case, the spectator ions are NH4+ and Cl.  Net ionic equation can be obtained from the complete ionic equation by removing the spectator ions.

6NH4+(aq)+2PO43(aq)+3Ca2+(aq)+6Cl(aq)Ca3(PO4)2(s)+6NH4+(aq)+6Cl(aq)

Thus net ionic equation can be written as follows:

    2PO43(aq)+3Ca2+(aq)Ca3(PO4)2(s)

(b)

Interpretation Introduction

Interpretation:

Concentration of spectator ions in the final solution with volume of 70.0mL obtained when 2.50g of (NH4)3PO4 is added to 50.0mL of 0.125MCaCl2.

(b)

Expert Solution
Check Mark

Answer to Problem I.26E

Concentration of ammonium ions is 0.72molL1 and chloride ions is 0.179molL1.

Explanation of Solution

Concentration of ammonium ions:

Number of moles of (NH4)3PO4 is calculated as shown below:

    n(NH4)3PO4=mM=2.50g149.09gmol1=0.0168mol

One mole of (NH4)3PO4 contains three moles of NH4+ ions.  Therefore, 0.0168mol of (NH4)3PO4 contains:

    nNH4+=0.0168mol(NH4)3PO4×6molNH4+ion2mol(NH4)3PO4=0.0504molNH4+

Therefore, the concentration of ammonium ions in 70.0mL of final solution can be calculated as follows:

    M=nV=0.0504mol0.07L=0.72molL1

Thus, the concentration of ammonium ions is 0.72molL1.

Concentration of chloride ions:

The concentration of the solution is 0.125M of CaCl2 and volume is 50.0mL.

One mole of CaCl2 contains two moles of Cl ions.  Therefore, 0.125M of CaCl2 contains 0.250M of Cl ions.  Given volume is 50.0mL.  Therefore, the moles of chloride ions in 50.0mL is calculated as follows:

    n=0.25molL1×0.0500L=0.0125mol

The number of moles of Cl ion is 0.0125mol and volume is 0.070L.  Therefore, the concentration of chloride ions can be calculated as follows:

    M=nV=0.0125mol0.07L=0.179molL1

Thus, the concentration of chloride ions is 0.179molL1.

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Chapter F Solutions

Chemical Principles: The Quest for Insight

Ch. F - Prob. A.1ECh. F - Prob. A.2ECh. F - Prob. A.3ECh. F - Prob. A.4ECh. F - Prob. A.5ECh. F - Prob. A.6ECh. F - Prob. A.7ECh. F - Prob. A.8ECh. F - Prob. A.9ECh. F - Prob. A.10ECh. F - Prob. A.11ECh. F - Prob. A.12ECh. F - Prob. A.13ECh. F - Prob. A.14ECh. F - Prob. A.15ECh. F - Prob. A.16ECh. F - Prob. A.17ECh. F - Prob. A.18ECh. F - Prob. A.19ECh. F - Prob. A.20ECh. F - Prob. A.21ECh. F - Prob. A.22ECh. F - Prob. A.23ECh. F - Prob. A.24ECh. F - Prob. A.25ECh. F - Prob. A.26ECh. F - Prob. A.27ECh. F - Prob. A.28ECh. F - Prob. A.29ECh. F - Prob. A.30ECh. F - Prob. A.31ECh. F - Prob. A.32ECh. F - Prob. A.33ECh. F - Prob. A.34ECh. F - Prob. A.35ECh. F - Prob. A.36ECh. F - Prob. A.37ECh. F - Prob. A.38ECh. F - Prob. A.39ECh. F - Prob. A.40ECh. F - Prob. A.41ECh. F - Prob. A.42ECh. F - Prob. B.1ASTCh. F - Prob. B.1BSTCh. F - Prob. B.2ASTCh. F - Prob. B.2BSTCh. F - Prob. B.3ASTCh. F - Prob. B.3BSTCh. F - Prob. B.1ECh. F - Prob. B.2ECh. F - Prob. B.3ECh. F - Prob. B.4ECh. F - Prob. 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